- #1
mhill
- 189
- 1
Hi, i am stuck at this problem , let be the divergent quantity
[tex] m= clog(\epsilon) +a_{0}+a_{1}g\epsilon ^{-1}+a_{2}g\epsilon ^{-2} +a_{3}g\epsilon ^{-3}+...+ [/tex]
where epsilon tends to 0 and g is just some coupling constant and c ,a_n are real numbers.
then i use the Borel transform of the function [tex] F(t)= \sum_{n=0}^{\infty}a_{n} \frac{t^{n}}{n!} [/tex] in this case
[tex] m= clog(\epsilon)+ \int_{0}^{\infty}dtF(t/\epsilon)e^{-t} [/tex]
my question is, does this last expression have only 2 divergent quantities ? , mainly the one due to log(e) and the second involving the poles of [tex] F(t/\epsilon) [/tex]
[tex] m= clog(\epsilon) +a_{0}+a_{1}g\epsilon ^{-1}+a_{2}g\epsilon ^{-2} +a_{3}g\epsilon ^{-3}+...+ [/tex]
where epsilon tends to 0 and g is just some coupling constant and c ,a_n are real numbers.
then i use the Borel transform of the function [tex] F(t)= \sum_{n=0}^{\infty}a_{n} \frac{t^{n}}{n!} [/tex] in this case
[tex] m= clog(\epsilon)+ \int_{0}^{\infty}dtF(t/\epsilon)e^{-t} [/tex]
my question is, does this last expression have only 2 divergent quantities ? , mainly the one due to log(e) and the second involving the poles of [tex] F(t/\epsilon) [/tex]